" b) "x(a-5)+y(5-a)

Find the equations of the tangent and the normal to the following curves : (a) 4x^(3)-3xy^(2)+6x^(2)-5xy-8y^(2)+9x+14=0 at the point (-2,3) (b) x^(5)+y^(5)-2xy=0 at the point (1,1)

A rectangle A B C D has its side A B parallel to line y=x , and vertices A ,Ba n dD lie on y=1,x=2, and x=-2, respectively. The locus of vertex C is x=5 (b) x-y=5 y=5 (d) x+y=5

A rectangle A B C D has its side A B parallel to line y=x , and vertices A ,Ba n dD lie on y=1,x=2, and x=-2, respectively. The locus of vertex C is x=5 (b) x-y=5 y=5 (d) x+y=5

A rectangle A B C D has its side A B parallel to line y=x , and vertices A ,Ba n dD lie on y=1,x=2, and x=-2, respectively. The locus of vertex C is x=5 (b) x-y=5 y=5 (d) x+y=5

A rectangle A B C D has its side A B parallel to line y=x , and vertices A ,Ba n dD lie on y=1,x=2, and x=-2, respectively. The locus of vertex C is (a) x=5 (b) x-y=5 (c) y=5 (d) x+y=5

x^(2)x2y^(3)x5x^(3)y^(2) is equal to: 10x^(2)y^(5) (b) 10x^(5)y^(2)10x^(5)y^(5)(d)x^(5)y^(5)

Factorise by taking out the common factors : 2x(a - b) + 3y(5a-5b) + 4z(2b - 2a)

The coordinates of the point on the curve y=6x-x^(2) the tangent at which is parallel to the line y=-4x are (a) (5,5) (b) (5,-5) (c) (-5,5) (d) (-5,-5)

The co-ordinate of the point on the curve y = 6x - x^2 , the tangent at which, is parallel to the line y =-4x are ....... A) (5,5) B) (5,-5) C) (-5,5) D) (-5,-5)

Simplify. (i) (x^2-5) (x+5)+25 (ii) (a^2+5) (b^3+3)+5 (iii) (t+s^2) (t^2-s) (iv) (a+b)(c-d)+(9a-b)(c+d)+2(ac+bd) (v) (x+y)(2x+y)+(x+2y)+(x+2y)(x-y) (vi) (x+y)(x^2-xy+y^2) (vii) (1.5x-4y)(1.5x+4y+3)-4.5x+12y (viii) (a+b+c)(a+b-c)